anonymous
  • anonymous
Let f(x,y)= 5/ (x^2 + y^2). Find an equation of a linear approximation to f(x,y) at (-1,2,1) and use it to estimate f(-1.05,2.1)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
wouldnt a linear approx just be a plane?
amistre64
  • amistre64
in fact, the plane is worked up similar to that of pont slope formula; except its point slope slope formula :)
amistre64
  • amistre64
\[z-z_o=f_x(x-x_o)+f_y(y-y_o)\]

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anonymous
  • anonymous
not sure @amistre64!
anonymous
  • anonymous
which one's x and whichone's x not etc.
amistre64
  • amistre64
\[f(x,y)=\frac{5}{x^2+y^2}\] \[f_x=\frac{-5(2x)}{(x^2+y^2)^2}\] \[f_y=\frac{-5(2y)}{(x^2+y^2)^2}\]
anonymous
  • anonymous
i'm sorry! can you possibly explain how you got fx and fy??
amistre64
  • amistre64
ok, let me post this and ill try given f( x=-1.05,y=2.1) that gives us: \[f_x=\frac{10(1.05)}{(1.05)^2+(2.1)^2}\] \[f_y=\frac{-10(2.1)}{(1.05)^2+(2.1)^2}\]
amistre64
  • amistre64
do you know partial derivatives?
anonymous
  • anonymous
yes
amistre64
  • amistre64
fx means we treat y as a constant \[\frac{5}{x^2+y^2}=5(x^2+c)^{-1}\] and derive it y is the same except for make x^2 = c
anonymous
  • anonymous
oh ok!
amistre64
  • amistre64
is that an OHHH ok? or one of those, i spose ... oks ?
amistre64
  • amistre64
in essense we ignore the variable that is not important in the partial and treat it as any other constant
amistre64
  • amistre64
they give us x0 and y0; to find z0 we solve f(x,y) for the given point and call f(x,y) = z
anonymous
  • anonymous
dat wuz an i kinda get it ok
amistre64
  • amistre64
if your confident with your usual derivative stuff; its just a little extra thinking to ignore that variable
amistre64
  • amistre64
f(x,y) = xy^2 fx = y^2 ; since Cx derives to C and C=y^2 Fy = 2xy since Cy^2 = 2Cy and C = x in that case
anonymous
  • anonymous
i get it!
amistre64
  • amistre64
lol, good :)
amistre64
  • amistre64
i knew this looked familiar
anonymous
  • anonymous
@amistre64 LOL Yup! i came in not that long ago, and saw the bump button available, so i pushed it! LOL
amistre64
  • amistre64
good thing it wasnt shiny and red :)
anonymous
  • anonymous
LOL:)

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